Two ports

310

Two-Port Networks

13.1 TERMINALS AND PORTS

In a two-terminal network, the terminal voltage is related to the terminal current by the impedanceZ ¼ V=I . In a four-terminal network, if each terminal pair (or port) is connected separately to anothercircuit as in Fig. 13-1, the four variables i1, i2, v1, and v2 are related by two equations called the terminalcharacteristics. These two equations, plus the terminal characteristics of the connected circuits, providethe necessary and sufficient number of equations to solve for the four variables.

13.2 Z-PARAMETERS

The terminal characteristics of a two-port network, having linear elements and dependent sources,may be written in the s-domain as

V1 ¼ Z11I1 þ Z12I2

V2 ¼ Z21I1 þ Z22I2ð1Þ

The coefficients Zij have the dimension of impedance and are called the Z-parameters of the network.The Z-parameters are also called open-circuit impedance parameters since they may be measured at oneterminal while the other terminal is open. They are

Z11 ¼V1

I1

��I2¼0

Z12 ¼V1

I2

��I1¼0

Z21 ¼V2

I1

��I2¼0

Z22 ¼V2

I2

��I1¼0

ð2Þ

Fig. 13-1

Copyright 2003, 1997, 1986, 1965 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

EXAMPLE 13.1 Find the Z-parameters of the two-port circuit in Fig. 13-2.

Apply KVL around the two loops in Fig. 13-2 with loop currents I1 and I2 to obtain

V1 ¼ 2I1 þ sðI1 þ I2Þ ¼ ð2þ sÞI1 þ sI2

V2 ¼ 3I2 þ sðI1 þ I2Þ ¼ sI1 þ ð3þ sÞI2ð3Þ

By comparing (1) and (3), the Z-parameters of the circuit are found to be

Z11 ¼ sþ 2

Z12 ¼ Z21 ¼ s

Z22 ¼ sþ 3

ð4Þ

Note that in this example Z12 ¼ Z21.

Reciprocal and Nonreciprocal Networks

A two-port network is called reciprocal if the open-circuit transfer impedances are equal;

Z12 ¼ Z21. Consequently, in a reciprocal two-port network with current I feeding one port, the

open-circuit voltage measured at the other port is the same, irrespective of the ports. The voltage is

equal to V ¼ Z12I ¼ Z21I. Networks containing resistors, inductors, and capacitors are generally

reciprocal. Networks that additionally have dependent sources are generally nonreciprocal (see

Example 13.2).

EXAMPLE 13.2 The two-port circuit shown in Fig. 13-3 contains a current-dependent voltage source. Find its

Z-parameters.

As in Example 13.1, we apply KVL around the two loops:

V1 ¼ 2I1 � I2 þ sðI1 þ I2Þ ¼ ð2þ sÞI1 þ ðs� 1ÞI2

V2 ¼ 3I2 þ sðI1 þ I2Þ ¼ sI1 þ ð3þ sÞI2

CHAP. 13] TWO-PORT NETWORKS 311

Fig. 13-2

Fig. 13-3

The Z-parameters are

Z11 ¼ sþ 2

Z12 ¼ s� 1

Z21 ¼ s

Z22 ¼ sþ 3

ð5Þ

With the dependent source in the circuit, Z12 6¼ Z21 and so the two-port circuit is nonreciprocal.

13.3 T-EQUIVALENT OF RECIPROCAL NETWORKS

A reciprocal network may be modeled by its T-equivalent as shown in the circuit of Fig. 13-4. Za,Zb, and Zc are obtained from the Z-parameters as follows.

Za ¼ Z11 � Z12

Zb ¼ Z22 � Z21

Zc ¼ Z12 ¼ Z21

ð6Þ

The T-equivalent network is not necessarily realizable.

EXAMPLE 13.3 Find the Z-parameters of Fig. 13-4.

Again we apply KVL to obtain

V1 ¼ ZaI1 þ ZcðI1 þ I2Þ ¼ ðZa þ ZcÞI1 þ ZcI2

V2 ¼ ZbI2 þ ZcðI1 þ I2Þ ¼ ZcI1 þ ðZb þ ZcÞI2ð7Þ

By comparing (1) and (7), the Z-parameters are found to be

Z11 ¼ Za þ Zc

Z12 ¼ Z21 ¼ Zc

Z22 ¼ Zb þ Zc

ð8Þ

13.4 Y-PARAMETERS

The terminal characteristics may also be written as in (9), where I1 and I2 are expressed in terms ofV1 and V2.

I1 ¼ Y11V1 þ Y12V2

I2 ¼ Y21V1 þ Y22V2

ð9Þ

The coefficients Yij have the dimension of admittance and are called the Y-parameters or short-circuitadmittance parameters because they may be measured at one port while the other port is short-circuited.The Y-parameters are

312 TWO-PORT NETWORKS [CHAP. 13

Fig. 13-4

Y11 ¼I1

V1

��V2¼0

Y12 ¼I1

V2

��V1¼0

Y21 ¼I2

V1

��V2¼0

Y22 ¼I2

V2

��V1¼0

ð10Þ

EXAMPLE 13.4 Find the Y-parameters of the circuit in Fig. 13-5.

We apply KCL to the input and output nodes (for convenience, we designate the admittances of the three

branches of the circuit by Ya, Yb, and Yc as shown in Fig. 13-6). Thus,

Ya ¼1

2þ 5s=3¼

3

5sþ 6

Yb ¼1

3þ 5s=2¼

2

5sþ 6

Yc ¼1

5þ 6=s¼

s

5sþ 6

ð11Þ

The node equations are

I1 ¼ V1Ya þ ðV1 � V2ÞYc ¼ ðYa þ YcÞV1 � YcV2

I2 ¼ V2Yb þ ðV2 � V1ÞYc ¼ �YcV1 þ ðYb þ YcÞV2

ð12Þ

By comparing (9) with (12), we get


Fig. 13-5

Fig. 13-6

Y11 ¼ Ya þ Yc

Y12 ¼ Y21 ¼ �Yc

Y22 ¼ Yb þ Yc

ð13Þ

Substituting Ya, Yb, and Yc in (11) into (13), we find

Y11 ¼sþ 3

5sþ 6

Y12 ¼ Y21 ¼�s

5sþ 6

Y22 ¼sþ 2

5sþ 6

ð14Þ

Since Y12 ¼ Y21, the two-port circuit is reciprocal.

13.5 PI-EQUIVALENT OF RECIPROCAL NETWORKS

A reciprocal network may be modeled by its Pi-equivalent as shown in Fig. 13-6. The threeelements of the Pi-equivalent network can be found by reverse solution. We first find the Y-parametersof Fig. 13-6. From (10) we have

Y11 ¼ Ya þ Yc [Fig. 13.7ðaÞ�

Y12 ¼ �Yc [Fig. 13-7ðbÞ�

Y21 ¼ �Yc [Fig. 13-7ðaÞ�

Y22 ¼ Yb þ Yc [Fig. 13-7ðbÞ�

ð15Þ

from which

Ya ¼ Y11 þ Y12 Yb ¼ Y22 þ Y12 Yc ¼ �Y12 ¼ �Y21 ð16Þ

The Pi-equivalent network is not necessarily realizable.

13.6 APPLICATION OF TERMINAL CHARACTERISTICS

The four terminal variables I1, I2, V1, and V2 in a two-port network are related by the two equations

(1) or (9). By connecting the two-port circuit to the outside as shown in Fig. 13-1, two additional

equations are obtained. The four equations then can determine I1, I2, V1, and V2 without any knowl-

edge of the inside structure of the circuit.


Fig. 13-7

EXAMPLE 13.5 The Z-parameters of a two-port network are given by

Z11 ¼ 2sþ 1=s Z12 ¼ Z21 ¼ 2s Z22 ¼ 2sþ 4

The network is connected to a source and a load as shown in Fig. 13-8. Find I1, I2, V1, and V2.

The terminal characteristics are given by

V1 ¼ ð2sþ 1=sÞI1 þ 2sI2

V2 ¼ 2sI1 þ ð2sþ 4ÞI2ð17Þ

The phasor representation of voltage vsðtÞ is Vs ¼ 12 V with s ¼ j. From KVL around the input and output loops

we obtain the two additional equations (18)

Vs ¼ 3I1 þ V1

0 ¼ ð1þ sÞI2 þ V2

ð18Þ

Substituting s ¼ j and Vs ¼ 12 in (17) and in (18) we get

V1 ¼ jI1 þ 2jI2

V2 ¼ 2jI1 þ ð4þ 2jÞI2

12 ¼ 3I1 þ V1

0 ¼ ð1þ jÞI2 þ V2

from which

I1 ¼ 3:29 �10:28 I2 ¼ 1:13 �131:28

V1 ¼ 2:88 37:58 V2 ¼ 1:6 93:88

13.7 CONVERSION BETWEEN Z- AND Y-PARAMETERS

The Y-parameters may be obtained from the Z-parameters by solving (1) for I1 and I2. ApplyingCramer’s rule to (1), we get

I1 ¼Z22

DZZ

V1 �Z12

DZZ

V2

I2 ¼�Z21

DZZ

V1 þZ11

DZZ

V2

ð19Þ

where DZZ ¼ Z11Z22 � Z12Z21 is the determinant of the coefficients in (1). By comparing (19) with (9)we have


Fig. 13-8

Y11 ¼Z22

DZZ

Y12 ¼�Z12

DZZ

Y21 ¼�Z21

DZZ

Y22 ¼Z11

DZZ

ð20Þ

Given the Z-parameters, for the Y-parameters to exist, the determinant DZZ must be nonzero. Con-versely, given the Y-parameters, the Z-parameters are

Z11 ¼Y22

DYY

Z12 ¼�Y12

DYY

Z21 ¼�Y21

DYY

Z22 ¼Y11

DYY

ð21Þ

where DYY ¼ Y11Y22 � Y12Y21 is the determinant of the coefficients in (9). For the Z-parameters of atwo-port circuit to be derived from its Y-parameters, DYY should be nonzero.

EXAMPLE 13.6 Referring to Example 13.4, find the Z-parameters of the circuit of Fig. 13-5 from its

Y-parameters.

The Y-parameters of the circuit were found to be [see (14)]

Y11 ¼sþ 3

5sþ 6Y12 ¼ Y21 ¼

�s

5sþ 6Y22 ¼

sþ 2

5sþ 6

Substituting into (21), where DYY ¼ 1=ð5sþ 6Þ, we obtain

Z11 ¼ sþ 2

Z12 ¼ Z21 ¼ s

Z22 ¼ sþ 3

ð22Þ

The Z-parameters in (22) are identical to the Z-parameters of the circuit of Fig. 13-2. The two circuits are

equivalent as far as the terminals are concerned. This was by design. Figure 13-2 is the T-equivalent of Fig. 13-5.

The equivalence between Fig. 13-2 and Fig. 13-5 may be verified directly by applying (6) to the Z-parameters given in

(22) to obtain its T-equivalent network.

13.8 h-PARAMETERS

Some two-port circuits or electronic devices are best characterized by the following terminalequations:

V1 ¼ h11I1 þ h12V2

I2 ¼ h21I1 þ h22V2

ð23Þ

where the hij coefficients are called the hybrid parameters, or h-parameters.

EXAMPLE 13.7 Find the h-parameters of Fig. 13-9.

This is the simple model of a bipolar junction transistor in its linear region of operation. By inspection, the

terminal characteristics of Fig. 13-9 are

V1 ¼ 50I1 and I2 ¼ 300I1 ð24Þ


By comparing (24) and (23) we get

h11 ¼ 50 h12 ¼ 0 h21 ¼ 300 h22 ¼ 0 ð25Þ

13.9 g-PARAMETERS

The terminal characteristics of a two-port circuit may also be described by still another set of hybridparameters given in (26).

I1 ¼ g11V1 þ g12I2

V2 ¼ g21V1 þ g22I2ð26Þ

where the coefficients gij are called inverse hybrid or g-parameters.

EXAMPLE 13.8 Find the g-parameters in the circuit shown in Fig. 13-10.

This is the simple model of a field effect transistor in its linear region of operation. To find the g-parameters,

we first derive the terminal equations by applying Kirchhoff’s laws at the terminals:

V1 ¼ 109I1At the input terminal:

V2 ¼ 10ðI2 � 10�3V1ÞAt the output terminal:

or I1 ¼ 10�9V1 and V2 ¼ 10I2 � 10�2

V1 (28)


g11 ¼ 10�9g12 ¼ 0 g21 ¼ �10�2

g22 ¼ 10 ð28Þ

13.10 TRANSMISSION PARAMETERS

The transmission parameters A, B, C, and D express the required source variables V1 and I1 in termsof the existing destination variables V2 and I2. They are called ABCD or T-parameters and are definedby


Fig. 13-9

Fig. 13-10

V1 ¼ AV2 � BI2

I1 ¼ CV2 �DI2ð29Þ

EXAMPLE 13.9 Find the T-parameters of Fig. 13-11 where Za and Zb are nonzero.

This is the simple lumped model of an incremental segment of a transmission line. From (29) we have

A ¼V1

V2

��I2¼0

¼Za þ Zb

Zb

¼ 1þ ZaYb

B ¼ �V1

I2

��V2¼0

¼ Za

C ¼I1

V2

��I2¼0

¼ Yb

D ¼ �I1

I2

��V2¼0

¼ 1

ð30Þ

13.11 INTERCONNECTING TWO-PORT NETWORKS

Two-port networks may be interconnected in various configurations, such as series, parallel, or

cascade connection, resulting in new two-port networks. For each configuration, certain set of

parameters may be more useful than others to describe the network.

Series Connection

Figure 13-12 shows a series connection of two two-port networks a and b with open-circuit

impedance parameters Za and Zb, respectively. In this configuration, we use the Z-parameters since

they are combined as a series connection of two impedances. The Z-parameters of the series connection

are (see Problem 13.10):


Fig. 13-11

Fig. 13-12

Z11 ¼ Z11;a þ Z11;b

Z12 ¼ Z12;a þ Z12;b

Z21 ¼ Z21;a þ Z21;b

Z22 ¼ Z22;a þ Z22;b

ð31aÞ

or, in the matrix form,

½Z� ¼ ½Za� þ ½Zb� ð31bÞ

Parallel Connection

Figure 13-13 shows a parallel connection of two-port networks a and b with short-circuit admittanceparameters Ya and Yb. In this case, the Y-parameters are convenient to work with. The Y-parametersof the parallel connection are (see Problem 13.11):

Y11 ¼ Y11;a þ Y11;b

Y12 ¼ Y12;a þ Y12;b

Y21 ¼ Y21;a þ Y21;b

Y22 ¼ Y22;a þ Y22;b

ð32aÞ

or, in the matrix form

½Y� ¼ ½Ya� þ ½Yb� ð32bÞ

Cascade Connection

The cascade connection of two-port networks a and b is shown in Fig. 13-14. In this case the

T-parameters are particularly convenient. The T-parameters of the cascade combination are

A ¼ AaAb þ BaCb

B ¼ AaBb þ BaDb

C ¼ CaAb þDaCb

D ¼ CaBb þDaDb

ð33aÞ

or, in the matrix form,

½T� ¼ ½Ta�½Tb� ð33bÞ


Fig. 13-13

13.12 CHOICE OF PARAMETER TYPE

What types of parameters are appropriate to and can best describe a given two-port network ordevice? Several factors influence the choice of parameters. (1) It is possible that some types ofparameters do not exist as they may not be defined at all (see Example 13.10). (2) Some parametersare more convenient to work with when the network is connected to other networks, as shown in Section13.11. In this regard, by converting the two-port network to its T- and Pi-equivalent and then applyingthe familiar analysis techniques, such as element reduction and current division, we can greatly reduceand simplify the overall circuit. (3) For some networks or devices, a certain type of parameter producesbetter computational accuracy and better sensitivity when used within the interconnected circuit.

EXAMPLE 13.10 Find the Z- and Y-parameters of Fig. 13-15.

We apply KVL to the input and output loops. Thus,

V1 ¼ 3I1 þ 3ðI1 þ I2ÞInput loop:

V2 ¼ 7I1 þ 2I2 þ 3ðI1 þ I2ÞOutput loop:

or V1 ¼ 6I1 þ 3I2 and V2 ¼ 10I1 þ 5I2 (34)


Z11 ¼ 6 Z12 ¼ 3 Z21 ¼ 10 Z22 ¼ 5

The Y-parameters are, however, not defined, since the application of the direct method of (10) or the conversion

from Z-parameters (19) produces DZZ ¼ 6ð5Þ � 3ð10Þ ¼ 0.

13.13 SUMMARY OF TERMINAL PARAMETERS AND CONVERSION

Terminal parameters are defined by the following equations

Z-parameters h-parameters T-parametersV1 ¼ Z11I1 þ Z12I2 V1 ¼ h11I1 þ h12V2 V1 ¼ AV2 � BI2V2 ¼ Z21I1 þ Z22I2 I2 ¼ h21I1 þ h22V2 I1 ¼ CV2 �DI2½V� ¼ ½Z�½I�

Y-parameters g-parametersI1 ¼ Y11V1 þ Y12V2 I1 ¼ g11V1 þ g12I2I2 ¼ Y21V1 þ Y22V2 V2 ¼ g21V1 þ g22I2½I� ¼ ½Y�½V�


Fig. 13-14

Fig. 13-15

Table 13-1 summarizes the conversion between the Z-, Y-, h-, g-, and T-parameters. For the

conversion to be possible, the determinant of the source parameters must be nonzero.

Solved Problems

13.1 Find the Z-parameters of the circuit in Fig. 13-16(a).

Z11 and Z21 are obtained by connecting a source to port #1 and leaving port #2 open [Fig. 13-16(b)].

The parallel and series combination of resistors produces

Z11 ¼V1

I1

��I2¼0

¼ 8 and Z21 ¼V2

I1

��I2¼0

¼1

3

Similarly, Z22 and Z12 are obtained by connecting a source to port #2 and leaving port #1 open [Fig.

13-16(c)].

Z22 ¼V2

I2

��I1¼0

¼8

9Z12 ¼

V1

I2

��I1¼0

¼1

3

The circuit is reciprocal, since Z12 ¼ Z21.


Table 13-1

Z Y h g T

Z

Z11 Z12 Y22

DYY

�Y12

DYY

Dhh

h22

h12

h22

1

g11

�g12

g11

A

C

DTT

C

Z21 Z22 �Y21

DYY

Y11

DYY

�h21

h22

1

h22

g21

g11

Dgg

g11

1

C

D

C

Y

Z22

Dzz

�Z12

Dzz

Y11 Y12 1

h11

�h12

h11

Dgg

g22

g12

g22

D

B

�DTT

B

�Z21

Dzz

Z11

Dzz

Y21 Y22 h21

h11

�Dnn

h11

�g21

g22

1

g22

�1

B

A

B

h

Dzz

Z22

Z12

Z22

1

Y11

�Y12

Y11

h11 h12g22

Dgg

g12

Dgg

B

D

DTT

D

�Z21

Z22

1

Z22

Y21

Y11

Dyy

Y11

h21 h22g21

Dgg

g11

Dgg

�1

D

C

D

g

1

Z11

�Z12

Z11

DYY

Y22

Y12

Y22

h22

Dhh

�h12

Dhh

g11 g12 C

A

�DTT

A

Z21

Z11

DZZ

Z11

�Y21

Y22

1

Y22

�h21

Dhh

h11

Dhh

g21 g22 1

A

B

A

T

Z11

Z21

DZZ

Z21

�Y22

Y21

�1

Y21

�Dhh

h21

�h11

h21

1

g21

g22

g21

A B

1

Z21

Z22

Z21

�DYY

Y21

�Y11

Y21

�h22

h21

�1

h21

g11

g21

Dgg

g21

C D

DPP ¼ P11P22 � P12P21 is the determinant of Z�; Y�; h�; g�; or T-parameters.

13.2 The Z-parameters of a two-port network N are given by

Z11 ¼ 2sþ 1=s Z12 ¼ Z21 ¼ 2s Z22 ¼ 2sþ 4

(a) Find the T-equivalent of N. (b) The network N is connected to a source and a load as shownin the circuit of Fig. 13-8. Replace N by its T-equivalent and then solve for i1, i2, v1, and v2.

(a) The three branches of the T-equivalent network (Fig. 13-4) are

Za ¼ Z11 � Z12 ¼ 2sþ1

s� 2s ¼

1

s

Zb ¼ Z22 � Z12 ¼ 2sþ 4� 2s ¼ 4

Zc ¼ Z12 ¼ Z21 ¼ 2s

(b) The T-equivalent of N, along with its input and output connections, is shown in phasor domain in Fig.

13-17.


Fig. 13-16

Fig. 13-17

By applying the familiar analysis techniques, including element reduction and current division, to

Fig. 13-17, we find i1, i2, v1, and v2.

In phasor domain In the time domain:

I1 ¼ 3:29 �10:28 i1 ¼ 3:29 cos ðt� 10:28ÞI2 ¼ 1:13 �131:28 i2 ¼ 1:13 cos ðt� 131:28ÞV1 ¼ 2:88 37:58 v1 ¼ 2:88 cos ðtþ 37:58ÞV2 ¼ 1:6 93:88 v2 ¼ 1:6 cos ðtþ 93:88Þ

13.3 Find the Z-parameters of the two-port network in Fig. 13-18.

KVL applied to the input and output ports obtains the following:

V1 ¼ 4I1 � 3I2 þ ðI1 þ I2Þ ¼ 5I1 � 2I2Input port:

V2 ¼ I2 þ ðI1 þ I2Þ ¼ I1 þ 2I2Output port:

By applying (2) to the above, Z11 ¼ 5, Z12 ¼ �2, Z21 ¼ 1, and Z22 ¼ 2:

13.4 Find the Z-parameters of the two-port network in Fig. 13-19 and compare the results with thoseof Problem 13.3.

KVL gives

V1 ¼ 5I1 � 2I2 and V2 ¼ I1 þ 2I2

The above equations are identical with the terminal characteristics obtained for the network of Fig.

13-18. Thus, the two networks are equivalent.

13.5 Find the Y-parameters of Fig. 13-19 using its Z-parameters.

From Problem 13.4,

Z11 ¼ 5; Z12 ¼ �2; Z21 ¼ 1; Z22 ¼ 2


Fig. 13-18

Fig. 13-19

Since DZZ ¼ Z11Z22 � Z12Z21 ¼ ð5Þð2Þ � ð�2Þð1Þ ¼ 12,

Y11 ¼Z22

DZZ

¼2

12¼

1

6Y12 ¼

�Z12

DZZ

¼2

12¼

1

6Y21 ¼

�Z21

DZZ

¼�1

12Y22 ¼

Z11

DZZ

¼5

12

13.6 Find the Y-parameters of the two-port network in Fig. 13-20 and thus show that the networks ofFigs. 13-19 and 13-20 are equivalent.

Apply KCL at the ports to obtain the terminal characteristics and Y-parameters. Thus,

I1 ¼V1

6þV2

6Input port:

I2 ¼V2

2:4�V1

12Output port:

Y11 ¼1

6Y12 ¼

1

6Y21 ¼

�1

12Y22 ¼

1

2:4¼

5

12and

which are identical with the Y-parameters obtained in Problem 3.5 for Fig. 13-19. Thus, the two networks

are equivalent.

13.7 Apply the short-circuit equations (10) to find the Y-parameters of the two-port network in Fig.13-21.

I1 ¼ Y11V1jV2¼0 ¼1

12þ

1

12

� �V1 or Y11 ¼

1

6

I1 ¼ Y12V2jV1¼0 ¼V2

4�V2

12¼

1

4�

1

12

� �V2 or Y12 ¼

1

6

I2 ¼ Y21V1jV2¼0 ¼ �V1

12or Y21 ¼ �

1

12

I2 ¼ Y22V2jV1¼0 ¼V2

3þV2

12¼

1

3þ

1

12

� �V2 or Y22 ¼

5

12


Fig. 13-20

Fig. 13-21

13.8 Apply KCL at the nodes of the network in Fig. 13-21 to obtain its terminal characteristics and Y-parameters. Show that two-port networks of Figs. 13-18 to 13-21 are all equivalent.

I1 ¼V1

12þV1 � V2

12þV2

4Input node:

I2 ¼V2

3þV2 � V1

12Output node:

I1 ¼1

6V1 þ

1

6V2 I2 ¼ �

1

12V1 þ

5

12V2

The Y-parameters observed from the above characteristic equations are identical with the Y-parameters of

the circuits in Figs. 13-18, 13-19, and 13-20. Therefore, the four circuits are equivalent.

13.9 Z-parameters of the two-port network N in Fig. 13-22(a) are Z11 ¼ 4s, Z12 ¼ Z21 ¼ 3s, andZ22 ¼ 9s. (a) Replace N by its T-equivalent. (b) Use part (a) to find input current i1 forvs ¼ cos 1000t (V).

(a) The network is reciprocal. Therefore, its T-equivalent exists. Its elements are found from (6) and

shown in the circuit of Fig. 13-22(b).


Fig. 13-22

Za ¼ Z11 � Z12 ¼ 4s� 3s ¼ s

Zb ¼ Z22 � Z21 ¼ 9s� 3s ¼ 6s

Zc ¼ Z12 ¼ Z21 ¼ 3s

(b) We repeatedly combine the series and parallel elements of Fig. 13-22(b), with resistors being in k� and s

in krad/s, to find Zin in k� as shown in the following.

ZinðsÞ ¼ Vs=I1 ¼ sþð3sþ 6Þð6sþ 12Þ

9sþ 18¼ 3sþ 4 or Zinð jÞ ¼ 3j þ 4 ¼ 5 36:98 k�

and i1 ¼ 0:2 cos ð1000t� 36:98Þ (mA).

13.10 Two two-port networks a and b, with open-circuit impedances Za and Zb, are connected in series(see Fig. 13-12). Derive the Z-parameters equations (31a).

From network a we have

V1a ¼ Z11;aI1a þ Z12;aI2a

V2a ¼ Z21;aI1a þ Z22;aI2a

From network b we have

V1b ¼ Z11;bI1b þ Z12;bI2b

V2b ¼ Z21;bI1b þ Z22;bI2b

From connection between a and b we have

I1 ¼ I1a ¼ I1b V1 ¼ V1a þ V1b

I2 ¼ I2a ¼ I2b V2 ¼ V2a þ V2b

Therefore,

V1 ¼ ðZ11;a þ Z11;bÞI1 þ ðZ12;a þ Z12;bÞI2

V2 ¼ ðZ21;a þ Z21;bÞI1 þ ðZ22;a þ Z22;bÞI2

from which the Z-parameters (31a) are derived.

13.11 Two two-port networks a and b, with short-circuit admittances Ya and Yb, are connected inparallel (see Fig. 13-13). Derive the Y-parameters equations (32a).

From network a we have

I1a ¼ Y11;aV1a þ Y12;aV2a

I2a ¼ Y21;aV1a þ Y22;aV2a

and from network b we have

I1b ¼ Y11;bV1b þ Y12;bV2b

I2b ¼ Y21;bV1b þ Y22;bV2b

From connection between a and b we have

V1 ¼ V1a ¼ V1b I1 ¼ I1a þ I1b

V2 ¼ V2a ¼ V2b I2 ¼ I2a þ I2b

Therefore,

I1 ¼ ðY11;a þ Y11;bÞV1 þ ðY12;a þ Y12;bÞV2

I2 ¼ ðY21;a þ Y21;bÞV1 þ ðY22;a þ Y22;bÞV2

from which the Y-parameters of (32a) result.


13.12 Find (a) the Z-parameters of the circuit of Fig. 13-23(a) and (b) an equivalent model which usesthree positive-valued resistors and one dependent voltage source.

(a) From application of KVL around the input and output loops we find, respectively,

V1 ¼ 2I1 � 2I2 þ 2ðI1 þ I2Þ ¼ 4I1

V2 ¼ 3I2 þ 2ðI1 þ I2Þ ¼ 2I1 þ 5I2

The Z-parameters are Z11 ¼ 4, Z12 ¼ 0, Z21 ¼ 2, and Z22 ¼ 5.

(b) The circuit of Fig. 13-23(b), with two resistors and a voltage source, has the same Z-parameters as the

circuit of Fig. 13-23(a). This can be verified by applying KVL to its input and output loops.

13.13 (a) Obtain the Y-parameters of the circuit in Fig. 13-23(a) from its Z-parameters. (b) Findan equivalent model which uses two positive-valued resistors and one dependent currentsource.

(a) From Problem 13.12, Z11 ¼ 4, Z12 ¼ 0, Z21 ¼ 2; Z22 ¼ 5, and so DZZ ¼ Z11Z22 � Z12Z21 ¼ 20.

Hence,

Y11 ¼Z22

DZZ

¼5

20¼

1

4Y12 ¼

�Z12

DZZ

¼ 0 Y21 ¼�Z21

DZZ

¼�2

20¼ �

1

10Y22 ¼

Z11

DZZ

¼4

20¼

1

5

(b) Figure 13-24, with two resistors and a current source, has the same Y-parameters as the circuit in Fig.

13-23(a). This can be verified by applying KCL to the input and output nodes.

13.14 Referring to the network of Fig. 13-23(b), convert the voltage source and its series resistor to itsNorton equivalent and show that the resulting network is identical with that in Fig. 13-24.

The Norton equivalent current source is IN ¼ 2I1=5 ¼ 0:4I1. But I1 ¼ V1=4. Therefore,

IN ¼ 0:4I1 ¼ 0:1V1. The 5-� resistor is then placed in parallel with IN . The circuit is shown in Fig.

13-25 which is the same as the circuit in Fig. 13-24.


Fig. 13-23

Fig. 13-24 Fig. 13-25

13.15 The h-parameters of a two-port network are given. Show that the network may be modeled bythe network in Fig. 13-26 where h11 is an impedance, h12 is a voltage gain, h21 is a current gain,and h22 is an admittance.

Apply KVL around the input loop to get

V1 ¼ h11I1 þ h12V2

Apply KCL at the output node to get

I2 ¼ h21I1 þ h22V2

These results agree with the definition of h-parameters given in (23).

13.16 Find the h-parameters of the circuit in Fig. 13-25.

By comparing the circuit in Fig. 13-25 with that in Fig. 13-26, we find

h11 ¼ 4 �; h12 ¼ 0; h21 ¼ �0:4; h22 ¼ 1=5 ¼ 0:2 ��1

13.17 Find the h-parameters of the circuit in Fig. 13-25 from its Z-parameters and compare with resultsof Problem 13.16.

Refer to Problem 13.13 for the values of the Z-parameters and DZZ. Use Table 13-1 for the conversion

of the Z-parameters to the h-parameters of the circuit. Thus,

h11 ¼DZZ

Z22

¼20

5¼ 4 h12 ¼

Z12

Z22

¼ 0 h21 ¼�Z21

Z22

¼�2

5¼ �0:4 h22 ¼

1

Z22

¼1

5¼ 0:2

The above results agree with the results of Problem 13.16.

13.18 The simplified model of a bipolar junction transistor for small signals is shown in the circuit ofFig. 13-27. Find its h-parameters.

The terminal equations are V1 ¼ 0 and I2 ¼ �I1. By comparing these equations with (23), we conclude

that h11 ¼ h12 ¼ h22 ¼ 0 and h21 ¼ �.


Fig. 13-26

Fig. 13-27

13.19 h-parameters of a two-port device H are given by

h11 ¼ 500 � h12 ¼ 10�4h21 ¼ 100 h22 ¼ 2ð10�6

Þ ��1

Draw a circuit model of the device made of two resistors and two dependent sources including thevalues of each element.

From comparison with Fig. 13-26, we draw the model of Fig. 13-28.

13.20 The device H of Problem 13-19 is placed in the circuit of Fig. 13-29(a). Replace H by its modelof Fig. 13-28 and find V2=Vs.


Fig. 13-28

Fig. 13-29

The circuit of Fig. 13-29(b) contains the model. With good approximation, we can reduce it to Fig.

13-29(c) from which

I1 ¼ Vs=2000 V2 ¼ �1000ð100I1Þ ¼ �1000ð100Vs=2000Þ ¼ �50Vs

Thus, V2=Vs ¼ �50.

13.21 A load ZL is connected to the output of a two-port device N (Fig. 13-30) whose terminalcharacteristics are given by V1 ¼ ð1=NÞV2 and I1 ¼ �NI2. Find (a) the T-parameters of Nand (b) the input impedance Zin ¼ V1=I1.

(a) The T-parameters are defined by [see (29)]V1 ¼ AV2 � BI2

I1 ¼ CV2 �DI2

The terminal characteristics of the device are

V1 ¼ ð1=NÞV2

I1 ¼ �NI2By comparing the two pairs of equations we get A ¼ 1=N, B ¼ 0, C ¼ 0, and D ¼ N.

(b) Three equations relating V1, I1, V2, and I2 are available: two equations are given by the terminal

characteristics of the device and the third equation comes from the connection to the load,

V2 ¼ �ZLI2

By eliminating V2 and I2 in these three equations, we get

V1 ¼ ZLI1=N2 from which Zin ¼ V1=I1 ¼ ZL=N

2

Supplementary Problems

13.22 The Z-parameters of the two-port network N in Fig. 13-22(a) are Z11 ¼ 4s, Z12 ¼ Z21 ¼ 3s, and Z22 ¼ 9s.

Find the input current i1 for vs ¼ cos 1000t (V) by using the open circuit impedance terminal characteristic

equations of N, together with KCL equations at nodes A, B, and C.

Ans: i1 ¼ 0:2 cos ð1000t� 36:98Þ (A)

13.23 Express the reciprocity criteria in terms of h-, g-, and T-parameters.

Ans: h12 þ h21 ¼ 0, g12 þ g21 ¼ 0, and AD� BC ¼ 1

13.24 Find the T-parameters of a two-port device whose Z-parameters are Z11 ¼ s, Z12 ¼ Z21 ¼ 10s, and

Z22 ¼ 100s. Ans: A ¼ 0:1;B ¼ 0;C ¼ 10�1=s, and D ¼ 10

13.25 Find the T-parameters of a two-port device whose Z-parameters are Z11 ¼ 106s, Z12 ¼ Z21 ¼ 107s, and

Z22 ¼ 108s. Compare with the results of Problem 13.21.


Fig. 13-30

Ans: A ¼ 0:1;B ¼ 0;C ¼ 10�7=s and D ¼ 10. For high frequencies, the device is similar to the device of

Problem 13.21, with N ¼ 10.

13.26 The Z-parameters of a two-port device N are Z11 ¼ ks, Z12 ¼ Z21 ¼ 10ks, and Z22 ¼ 100ks. A 1-� resistor

is connected across the output port (Fig. 13-30). (a) Find the input impedance Zin ¼ V1=I1 and construct

its equivalent circuit. (b) Give the values of the elements for k ¼ 1 and 106.

Ans: ðaÞ Zin ¼ks

1þ 100ks¼

1

100þ 1=ks

The equivalent circuit is a parallel RL circuit with R ¼ 10�2 � and L ¼ 1 kH:

ðbÞ For k ¼ 1;R ¼1

100� and L ¼ 1 H. For k ¼ 106;R ¼

1

100� and L ¼ 106 H

13.27 The device N in Fig. 13-30 is specified by its following Z-parameters: Z22 ¼ N2Z11 and

Z12 ¼ Z21 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZ11Z22

p¼ NZ11. Find Zin ¼ V1=I1 when a load ZL is connected to the output terminal.

Show that if Z11 � ZL=N2 we have impedance scaling such that Zin ¼ ZL=N

2.

Ans: Zin ¼ZL

N2 þ ZL=Z11

. For ZL � N2Z11;Zin ¼ ZL=N

2

13.28 Find the Z-parameters in the circuit of Fig. 13-31. Hint: Use the series connection rule.

Ans: Z11 ¼ Z22 ¼ sþ 3þ 1=s;Z12 ¼ Z21 ¼ sþ 1

13.29 Find the Y-parameters in the circuit of Fig. 13-32. Hint: Use the parallel connection rule.

Ans: Y11 ¼ Y22 ¼ 9ðsþ 2Þ=16;Y12 ¼ Y21 ¼ �3ðsþ 2Þ=16


Fig. 13-31

Fig. 13-32

13.30 Two two-port networks a and b with transmission parameters Ta and Tb are connected in cascade (Fig. 13-

14). Given I2a ¼ �I1b and V2a ¼ V1b, find the T-parameters of the resulting two-port network.

Ans: A ¼ AaAb þ BaCb, B ¼ AaBb þ BaDb, C ¼ CaAb þDaCb, D ¼ CaBb þDaDb

13.31 Find the T- and Z-parameters of the network in Fig. 13-33. The impedances of capacitors are given. Hint:

Use the cascade connection rule.

Ans: A ¼ 5j � 4, B ¼ 4j þ 2, C ¼ 2j � 4, and D ¼ j3, Z11 ¼ 1:3� 0:6j, Z22 ¼ 0:3� 0:6j,Z12 ¼ Z21 ¼ �0:2� 0:1j

13.32 Find the Z-parameters of the two-port circuit of Fig. 13-34.

Ans: Z11 ¼ Z22 ¼12ðZb þ ZaÞ;Z12 ¼ Z21 ¼

12ðZb � ZaÞ

13.33 Find the Z-parameters of the two-port circuit of Fig. 13-35.

Ans: Z11 ¼ Z22 ¼1

2

Zbð2Za þ ZbÞ

Za þ Zb

; Z12 ¼ Z21 ¼1

2

Z2b

Za þ Zb

13.34 Referring to the two-port circuit of Fig. 13-36, find the T-parameters as a function of ! and specify their

values at ! ¼ 1, 103, and 106 rad/s.


Fig. 13-33

Fig. 13-34

Fig. 13-35

Ans: A ¼ 1� 10�9!2þ j10�9!, B ¼ 10�3

ð1þ j!Þ, C ¼ 10�6j!, and D ¼ 1. At ! ¼ 1 rad/s, A ¼ 1,

B ¼ 10�3ð1þ jÞ, C ¼ 10�6j, and D ¼ 1. At ! ¼ 103 rad/s, A � 1, B � j, C ¼ 10�3j, and D ¼ 1.

At ! ¼ 106 rad/s, A � �103, B � 103j, C ¼ j, and D ¼ 1

13.35 A two-port network contains resistors, capacitors, and inductors only. With port #2 open [Fig. 13-37(a)], a

unit step voltage v1 ¼ uðtÞ produces i1 ¼ e�tuðtÞ ðmAÞ and v2 ¼ ð1� e�tÞuðtÞ (V). With port #2 short-

circuited [Fig. 13-37(b)], a unit step voltage v1 ¼ uðtÞ delivers a current i1 ¼ 0:5ð1þ e�2tÞuðtÞ ðmAÞ. Find

i2 and the T-equivalent network. Ans: i2 ¼ 0:5ð�1þ e�2tÞuðtÞ [see Fig. 13-37(c)]

13.36 The two-port network N in Fig. 13-38 is specified by Z11 ¼ 2, Z12 ¼ Z21 ¼ 1, and Z22 ¼ 4. Find I1, I2, and

I3. Ans: I1 ¼ 24 A; I2 ¼ 1:5 A; and I3 ¼ 6:5 A


Fig. 13-37

Fig. 13-38

Fig. 13-36

Engineering

Two ports